Question: Evaluate the double integral. $ \int_{-1}^1 \int_0^{x^2} x + 3y \, dy \, dx =$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{5}$ (Choice B) B $\dfrac{4}{5}$ (Choice C) C $\dfrac{2}{5}$ (Choice D) D $\dfrac{3}{5}$
Answer: First, we evaluate the inner integral. We can substitute in the $x^2$ at the end as if it were a numerical bound. $\begin{aligned} \int_{-1}^1 \int_0^{x^2} x + 3y \, dy \, dx &= \int_{-1}^1 \left[ xy + \dfrac{3y^2}{2} \right]_0^{x^2} \, dx \\ \\ &= \int_{-1}^1 x^3 + \dfrac{3x^4}{2} \, dx \end{aligned}$ Second, we evaluate the outer integral. $\begin{aligned} \int_{-1}^1 x^3 + \dfrac{3x^4}{2} \, dx &= \left[ \dfrac{x^4}{4} + \dfrac{3x^5}{10} \right]_{-1}^1 \\ \\ &= \dfrac{1}{4} + \dfrac{3}{10} - \left( \dfrac{1}{4} - \dfrac{3}{10} \right) \\ \\ &= \dfrac{3}{5} \end{aligned}$ The answer: $ \int_{-1}^1 \int_0^{x^2} x + 3y \, dy \, dx = \dfrac{3}{5}$